Question

An LED is placed at a depth of h below the water surface. An opaque disc is floating on the surface of the water such that the bulb is not visible from the surface. The minimum radius of the disc will be ______.

Options

• `(2h)/mu`

• `h/sqrt(mu - 1)`

• `h/(2mu - 1)`

• `mu/(2h)`

Answer 1

An LED is placed at a depth of h below the water surface. An opaque disc is floating on the surface of the water such that the bulb is not visible from the surface. The minimum radius of the disc will be `underlinebb(h/sqrt(mu - 1))`.

Explanation:

The given situation is shown below.

The light from the LED will not emerge from the water if the incidence angle is greater than the critical angle near the edge of the disc, as shown in the image.

i.e., i > C

or sini > sinC ..............(i)

Now, if R is the radius of the disc and λ is the depth of the LED, then

`sini = R/sqrt(R^2 + h^2)` and `sinC = 1/mu`

From Eq (i), we have

`R/(sqrt(R^2 + h^2)) > 1/mu ⇒ R > h/sqrt(mu - 1)`